How do you describe the end behavior of $f (x) = x^{2} - 8 x + 18$ $f(x)=x^2-8x+18$ ?

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How do you describe the end behavior of $f (x) = x^{2} - 8 x + 18$ $f(x)=x^2-8x+18$ ?

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Answer 1 · 2016-09-29T02:16:26

as $x \to - \infty, f (x) \to + \infty$ $xrarr-oo, f(x)rarr+oo$ and
as $x \to + \infty, f (x) \to + \infty$ $xrarr+oo, f(x)rarr+oo$

Explanation:

$f (x) = 1 x^{2} - 8 x + 18$ $f(x)=color(red)1x^color(blue)2-8x+18$

Because the degree $2$ $color(blue)2$ is even, this an even function. Even functions have end behaviors that both go in the same direction in y.

The function has a positive leading coefficient, $1$ $color(red)1$ . Even functions with positive leading coefficients have end behaviors that both go toward positive infinity (both ends of this quadratic/parabola graph point "up").

In other words,

as $x \to - \infty, f (x) \to + \infty$ $xrarr-oo, f(x)rarr+oo$ and
as $x \to + \infty, f (x) \to + \infty$ $xrarr+oo, f(x)rarr+oo$

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How do you describe the end behavior of $f (x) = x^{2} - 8 x + 18$ $f(x)=x^2-8x+18$ ?

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Explanation:

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How do you describe the end behavior of f(x)=x^2-8x+18f(x)=x2−8x+18f(x)=x^2-8x+18?

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Explanation:

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How do you describe the end behavior of $f (x) = x^{2} - 8 x + 18$ $f(x)=x^2-8x+18$ ?